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    domain of f

    f(x) = (square root -x) / x^3 + 2x^2 - 3x find the domain of f
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    Quote Originally Posted by robinpatrick View Post
    f(x) = (square root -x) / x^3 + 2x^2 - 3x find the domain of f
    1)The square root needs to be non-negative.
    Thus, \sqrt{-x} for -x\geq 0 thus, x\leq 0.

    2)The numerator cannot be zero.
    x^3+2x^2-3x=0
    x(x^2+2x-3)=0
    x(x+3)(x-1)=0
    Thus,
    x\not = 0,1,-3.

    Since x\leq 0 the only one of those that works is x=-3 thus,
    x\leq 0 \mbox{ and } x\not = -3
    Thus,
    (-\infty,-3)\cup (-3,0]
    -----
    [Note for you, for anyone interested in anything I have to say].
    These domain seem not be be well-defined, because to definied a function you need to create a domain and co-domain which you did not do here. Thus, how can you find a domain?
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    Quote Originally Posted by ThePerfectHacker View Post
    1)The square root needs to be non-negative.
    Thus, \sqrt{-x} for -x\geq 0 thus, x\leq 0.

    2)The numerator cannot be zero.
    x^3+2x^2-3x=0
    x(x^2+2x-3)=0
    x(x+3)(x-1)=0
    Thus,
    x\not = 0,1,-3.

    Since x\leq 0 the only one of those that works is x=-3 thus,
    x\leq 0 \mbox{ and } x\not = -3
    Thus,
    (-\infty,-3)\cup (-3,0]
    -----
    [Note for you, for anyone interested in anything I have to say].
    These domain seem not be be well-defined, because to definied a function you need to create a domain and co-domain which you did not do here. Thus, how can you find a domain?
    Slight correction. x cannot be 0 else the denominator is zero. So the domain would be:

    (-\infty,-3)\cup (-3,0)

    -Dan
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